What is the area with respect to the y axis of the curve x2 = 3y, from x = 0 to x = 4?
Let us find the area "A" with respect to the y-axis of the curve "x^2 = 3y," from "x = 0" to "x = 4." If "x=0"then "y=0." If "x=4" then "y=\\frac{16}3." It follows that "x=\\pm\\sqrt{3y}," and using the symmetry of the curve about the y-axis, we get that
"A=2\\int\\limits_0^{\\frac{16}3}\\sqrt{3y}dy\n=2\\sqrt{3}\\frac{2}3y^{\\frac{3}2}|_0^{\\frac{16}3}\n=\\frac{4}3\\sqrt{3}(\\frac{16}3)^{\\frac{3}2}\n=\\frac{4}3\\sqrt{3}\\frac{4^3}{3\\sqrt{3}}\n=\\frac{256}{9}" (sq. units).
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